Are these rational curves? I have to find the singular points of the following curves and tell if they are rational. The curves are $C=Z(x^2+y^2+x^2y^2)$ and $C=Z(x^3+y^3-1)$, and the base field is the complex one.
I think I found the singular points but I'm stuck in finding the rational morphism. Can some one help me?  
 A: A plane  affine curve is  rational iff its projectivization is rational, so let's study these projectivizations since we have more tools at our disposal in the projective plane.  
2) The projectivization of the second curve is the curve $x^3+y^3-z^3=0$.
This is a smooth curve of genus $\frac {(3-1)(3-2)}{2}=1$, so it is not rational since rational curves have genus $0$ .
1)  The projectivization of the first  curve is the curve $C$ with equation $x^2z^2+y^2z^2+x^2y^2=0$, which has three nodal singularities at $(1:0:0),(0:1:0),(0:0:1)$.
The birational transformation of the plane (traditionally "quadratic transformation") $x=\frac {1}{X},  y=\frac {1}{Y},   z=\frac {1}{Z}$ transforms (after multiplying by $X^2Y^2Z^2$) the curve  $C$ into the conic $C'$ of equation $Y^2+X^2+Z^2=0$.
So the curve $C$ is indeed rational, since the conic $C'$ is rational .
Another argument for the  rationality of $C$ is that its geometric genus is $\frac {(4-1)(4-2)}{2}-3\cdot 1=0$ and  a curve of geometric genus $0$ is rational (the subtracted term in the formula comes from the the three nodal singularities of the curve).   
